# Check if a string is made up of K alternating characters

Given a string str and an integer K, the task is to check if it is made up of K alternating characters.

Examples:

Input: str = “acdeac”, K = 4
Output: Yes

Input: str = “abcdcab”, K = 2
Output: No

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: In order for the string to be made up of K alternating characters, it must satisfy the following conditions:

1. All the characters at (i + mK) indices must be same, where i is the current index and mK represents the mth multiple of K. It means that after every K indices, the character must get repeated.
2. The adjacent characters must not be the same. This is because if the string is of type “AAAAA”, where a single character is repeated any number of time, the above condition will get matched, but the answer must be ‘No’ in this case.

Below code is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to check if a string ` `// is made up of k alternating characters ` `bool` `isKAlternating(string s, ``int` `k) ` `{ ` `    ``if` `(s.length() < k) ` `        ``return` `false``; ` ` `  `    ``int` `checker = 0; ` ` `  `    ``// Check if all the characters at ` `    ``// indices 0 to K-1 are different ` `    ``for` `(``int` `i = 0; i < k; i++) { ` ` `  `        ``int` `bitAtIndex = s[i] - ``'a'``; ` ` `  `        ``// If that bit is already set in ` `        ``// checker, return false ` `        ``if` `((checker & (1 << bitAtIndex)) > 0) { ` `            ``return` `false``; ` `        ``} ` ` `  `        ``// Otherwise update and continue by ` `        ``// setting that bit in the checker ` `        ``checker = checker | (1 << bitAtIndex); ` `    ``} ` ` `  `    ``for` `(``int` `i = k; i < s.length(); i++) ` `        ``if` `(s[i - k] != s[i]) ` `            ``return` `false``; ` ` `  `    ``return` `true``; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string str = ``"acdeac"``; ` `    ``int` `K = 4; ` ` `  `    ``if` `(isKAlternating(str, K)) ` `        ``cout << ``"Yes"` `<< endl; ` `    ``else` `        ``cout << ``"No"` `<< endl; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG{ ` `  `  `// Function to check if a String ` `// is made up of k alternating characters ` `static` `boolean` `isKAlternating(String s, ``int` `k) ` `{ ` `    ``if` `(s.length() < k) ` `        ``return` `false``; ` `  `  `    ``int` `checker = ``0``; ` `  `  `    ``// Check if all the characters at ` `    ``// indices 0 to K-1 are different ` `    ``for` `(``int` `i = ``0``; i < k; i++) { ` `  `  `        ``int` `bitAtIndex = s.charAt(i) - ``'a'``; ` `  `  `        ``// If that bit is already set in ` `        ``// checker, return false ` `        ``if` `((checker & (``1` `<< bitAtIndex)) > ``0``) { ` `            ``return` `false``; ` `        ``} ` `  `  `        ``// Otherwise update and continue by ` `        ``// setting that bit in the checker ` `        ``checker = checker | (``1` `<< bitAtIndex); ` `    ``} ` `  `  `    ``for` `(``int` `i = k; i < s.length(); i++) ` `        ``if` `(s.charAt(i - k) != s.charAt(i) ) ` `            ``return` `false``; ` `  `  `    ``return` `true``; ` `} ` `  `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``String str = ``"acdeac"``; ` `    ``int` `K = ``4``; ` `  `  `    ``if` `(isKAlternating(str, K)) ` `        ``System.out.print(``"Yes"` `+``"\n"``); ` `    ``else` `        ``System.out.print(``"No"` `+``"\n"``);  ` `} ` `} ` ` `  `// This code is contributed by sapnasingh4991 `

## Python3

 `# Python 3 implementation of the approach ` `  `  `# Function to check if a string ` `# is made up of k alternating characters ` `def` `isKAlternating( s, k): ` `    ``if` `(``len``(s) < k): ` `        ``return` `False` `  `  `    ``checker ``=` `0` `  `  `    ``# Check if all the characters at ` `    ``# indices 0 to K-1 are different ` `    ``for` `i ``in` `range``( k): ` `  `  `        ``bitAtIndex ``=` `ord``(s[i]) ``-` `ord``(``'a'``) ` `  `  `        ``# If that bit is already set in ` `        ``# checker, return false ` `        ``if` `((checker & (``1` `<< bitAtIndex)) > ``0``): ` `            ``return` `False` `  `  `        ``# Otherwise update and continue by ` `        ``# setting that bit in the checker ` `        ``checker ``=` `checker | (``1` `<< bitAtIndex) ` `  `  `    ``for` `i ``in` `range``(k,``len``(s)): ` `        ``if` `(s[i ``-` `k] !``=` `s[i]): ` `            ``return` `False` `  `  `    ``return` `True` `  `  `# Driver code ` `if` `__name__ ``=``=``"__main__"``: ` ` `  `    ``st ``=` `"acdeac"` `    ``K ``=` `4` `  `  `    ``if` `(isKAlternating(st, K)): ` `        ``print` `(``"Yes"``) ` `    ``else``: ` `        ``print` `(``"No"``) ` `  `  `# This code is contributed by chitranayal    `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG{ ` ` `  `// Function to check if a String ` `// is made up of k alternating characters ` `static` `bool` `isKAlternating(String s, ``int` `k) ` `{ ` `    ``if` `(s.Length < k) ` `        ``return` `false``; ` ` `  `    ``int` `checker = 0; ` ` `  `    ``// Check if all the characters at ` `    ``// indices 0 to K-1 are different ` `    ``for` `(``int` `i = 0; i < k; i++) { ` ` `  `        ``int` `bitAtIndex = s[i] - ``'a'``; ` ` `  `        ``// If that bit is already set in ` `        ``// checker, return false ` `        ``if` `((checker & (1 << bitAtIndex)) > 0) { ` `            ``return` `false``; ` `        ``} ` ` `  `        ``// Otherwise update and continue by ` `        ``// setting that bit in the checker ` `        ``checker = checker | (1 << bitAtIndex); ` `    ``} ` ` `  `    ``for` `(``int` `i = k; i < s.Length; i++) ` `        ``if` `(s[i - k] != s[i] ) ` `            ``return` `false``; ` ` `  `    ``return` `true``; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main()  ` `{ ` `    ``String str = ``"acdeac"``; ` `    ``int` `K = 4; ` ` `  `    ``if` `(isKAlternating(str, K)) ` `        ``Console.WriteLine(``"Yes"``); ` `    ``else` `        ``Console.WriteLine(``"No"``); ` `} ` `} ` ` `  `// This article contributed by AbhiThakur `

Output:

```Yes
```

Time Complexity: O(N)

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